Answer
Yes, this matrix is invertible.
Work Step by Step
By Theorem 4, $\det A=0$ if and only if $A$ is not invertible, so we compute the determinant in order to decide whether the given matrix is invertible.
$\begin{vmatrix}2&0&0&6\\1&-7&-5&0\\3&8&6&0\\0&7&5&4\end{vmatrix}=2\begin{vmatrix}-7&-5&0\\8&6&0\\7&5&4\end{vmatrix}-6\begin{vmatrix}1&-7&-5\\3&8&6\\0&7&5\end{vmatrix}=2\cdot 4\begin{vmatrix}-7&-5\\8&6\end{vmatrix}-6\cdot \left (1\begin{vmatrix}8&6\\7&5\end{vmatrix}-3\begin{vmatrix}-7&-5\\5&7\end{vmatrix} \right )=8\cdot (-2)-6 \cdot (-2)+18\cdot0=-4$
We see that the determinant is nonzero, so the matrix is invertible.
